Physics 2024

Q2

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As shown in the figure, a $6 m$ long and $3 m$ wide boat moves along a narrow passageway of water next to a granite cliff. The boat is moving at a steady speed of $15 m/s$ parallel to the cliff at a constant distance of $10 m$ away (measured from center of the boat shown in the figure). As shown, a small piece of rock slides off the edge of a 500 cliff at $2 m/s$ when the front tip of the boat is $25 m$ away from the rock. The rock falls from a height $20 m$ above the tip of the boat.

a) We denote the time for the rock to reach the same level as the front tip of the boat as $t_{rock}$ and the time for the boat to reach where the rock has fallen as $t_{boat}$ . By calculating $t_{rock}$ , $t_{boat}$ and doing any other relevant calculations, determine (with a short explanation) if the rock will hit the boat. Will the rock hit the boat?

b) After the rock slides off at $50^{\circ}$ below the horizontal from the cliff, how will the angle of the velocity of the rock be subsequently?
a. Increasing to more than $50^{\circ}$ below the horizontal.
b. Stay constant at $50^{\circ}$ below the horizontal.
c. Reducing to less than $50^{\circ}$ below the horizontal.

ANSWER

Good evening! Welcome to our virtual physics session. That's an excellent problem you've brought in—it combines constant velocity motion with projectile motion, which is a classic scenario. Let's break it down together, step-by-step, to make sure every part is clear.

Part a) Will the rock hit the boat?

To solve this, we need to answer a fundamental question: Are the rock and the boat in the same place at the same time? This means we need to analyze two things:

Timing: Does the rock reach the water level during the time interval that the boat is passing underneath?
Position: If the timing is right, does the rock land within the boat's width?

Let's start by calculating the times you mentioned, $t_{rock}$ and $t_{boat}$ . We'll consider the acceleration due to gravity, $g$ , to be $9.81 {m/s}^{2}$ .

Step 1: Calculate the time for the rock to fall ( $t_{rock}$ )

The rock's fall is a projectile motion problem. The most effective way to handle these is to separate the motion into horizontal (x) and vertical (y) components.

Let's define our coordinate system:

The origin (0, 0) is at the water level directly below where the rock starts to fall.
The positive y-direction is upwards.
The positive x-direction is away from the cliff.

The rock starts at an initial height $h = 20 m$ . It falls to the water level, so its vertical displacement is $Δ y = - 20 m$ .

The initial velocity is $v_{0} = 2 m/s$ at $50^{\circ}$ below the horizontal. Let's find its vertical component, $v_{0 y}$ :
$v_{0 y} = - v_{0} \sin (50^{\circ}) = - 2 \sin (50^{\circ}) \approx - 1.532 m/s$
The negative sign is crucial because it's directed downwards.

Now, we can use the kinematic equation for vertical motion:

Δ y = v_{0 y} t + \frac{1}{2} a_{y} t^{2}

Plugging in our values:

- 20 = (- 1.532) t_{rock} + \frac{1}{2} (- 9.81) t_{rock}^{2}

- 20 = - 1.532 t_{rock} - 4.905 t_{rock}^{2}

Rearranging this gives us a quadratic equation:

4.905 t_{rock}^{2} + 1.532 t_{rock} - 20 = 0

We can solve for $t_{rock}$ using the quadratic formula, $t = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ . We'll take the positive root, as time cannot be negative.

t_{rock} = \frac{- 1.532 + \sqrt{(1.532)^{2} - 4 (4.905) (- 20)}}{2 (4.905)}

t_{rock} = \frac{- 1.532 + \sqrt{2.347 + 392.4}}{9.81}

t_{rock} = \frac{- 1.532 + 19.868}{9.81} \approx 1.869 s

So, the rock takes approximately $t_{rock} = 1.87 s$ to hit the water.

Step 2: Calculate the time for the boat to reach the impact zone ( $t_{boat}$ )

The boat moves at a constant speed, so this calculation is more straightforward.

Distance for the front tip of the boat to travel, $d = 25 m$ .
Speed of the boat, $v = 15 m/s$ .

The time it takes for the front of the boat to arrive at the point directly below the rock is:

t_{boat} = \frac{d}{v} = \frac{25 m}{15 m/s} = \frac{5}{3} \approx 1.667 s

So, the front tip of the boat reaches the "danger zone" at $t_{boat} = 1.67 s$ .

Step 3: Compare times and check the boat's position

The rock hits the water at $t = 1.87 s$ . The front of the boat passes this point at $t = 1.67 s$ . This means the front of the boat has already passed the impact point before the rock lands.

But is the rest of the boat clear? Let's find out where the back of the boat is when the rock lands. The boat is $6 m$ long.

The time window during which the boat is passing under the rock's fall path is from when the front tip arrives ( $1.67 s$ ) to when the rear tip passes.
Time for the rear to pass = Time for front to pass + Time to travel boat's length

t_{rear} = t_{boat} + \frac{{length}_{boat}}{v_{boat}} = 1.67 s + \frac{6 m}{15 m/s} = 1.67 s + 0.4 s = 2.07 s

So, some part of the boat is in the danger zone between $t = 1.67 s$ and $t = 2.07 s$ .
The rock lands at $t = 1.87 s$ , which is within this time window! So, it's possible it hits the boat. Now we need to check the horizontal position.

Step 4: Check the rock's horizontal landing position

Does the rock actually land on the boat's deck? Let's calculate how far the rock travels horizontally away from the cliff.
The initial horizontal velocity is:
$v_{0 x} = v_{0} \cos (50^{\circ}) = 2 \cos (50^{\circ}) \approx 1.286 m/s$

Ignoring air resistance, there's no horizontal acceleration. The horizontal distance ( $Δ x$ ) it travels is:

Δ x = v_{0 x} \times t_{rock} = 1.286 m/s \times 1.869 s \approx 2.40 m

The rock lands $2.40 m$ away from the cliff face.

Now, where is the boat?

The center of the boat is $10 m$ from the cliff.
The boat is $3 m$ wide.
This means the boat occupies the space from $10 - 1.5 = 8.5 m$ to $10 + 1.5 = 11.5 m$ away from the cliff.

The rock lands at $2.40 m$ from the cliff, while the boat is between $8.5 m$ and $11.5 m$ .

Conclusion for Part (a)

No, the rock will not hit the boat.

Explanation: While our time calculation shows that the boat is indeed passing under the drop point when the rock reaches the water level (the rock lands at $1.87 s$ , and the boat is passing underneath between $1.67 s$ and $2.07 s$ ), the rock's horizontal travel is insufficient. The rock only travels $2.40 m$ horizontally from the cliff, landing in the water well short of the boat, which is located between $8.5 m$ and $11.5 m$ from the cliff.

Part b) The angle of the rock's velocity

This is an excellent conceptual question about the nature of projectile motion. Let's analyze the velocity components.

Horizontal Velocity ( $v_{x}$ ): Since we ignore air resistance, there are no horizontal forces acting on the rock. According to Newton's first law, its horizontal velocity will remain constant throughout the fall. $v_{x} = constant$ .
Vertical Velocity ( $v_{y}$ ): The rock is constantly being accelerated downwards by gravity. This means its downward speed continuously increases. $| v_{y} |$ increases with time.

The angle of the velocity vector below the horizontal, $θ$ , is related to these components by:

\tan (θ) = \frac{vertical component}{horizontal component} = \frac{| v_{y} |}{| v_{x} |}

Since $| v_{y} |$ is increasing and $| v_{x} |$ is constant, the ratio $\frac{| v_{y} |}{| v_{x} |}$ must be continuously increasing as the rock falls.
If $\tan (θ)$ is increasing, then the angle $θ$ itself must be increasing.

Think of it this way: gravity is always pulling the rock more and more vertically. So, its path will become steeper and steeper. It starts at $50^{\circ}$ and will only get steeper from there.

Therefore, the correct answer is:
a. Increasing to more than $50^{\circ}$ below the horizontal.

Q3

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In the figure shown, two blocks are connected by a massless cable round a smooth pulley where friction is negligible. The block on the $30^{\circ}$ slope has mass $2 M$ and the other suspended block has mass $M$ . The blocks are moving downwards slowly and the coefficient of kinetic friction between the block with mass $2 M$ and the incline is $0.20$ .

Determine the magnitude of the acceleration of the suspended block as it falls (before the block of mass $2 M$ moving down the incline hits the pulley).
On the axes below, sketch to show how the distance of the suspended block (of mass $M$ ) from the pulley changes with time (before the block of mass $2 M$ moving down the incline hits the pulley).
Suggest one modification that can be made so that the magnitude of the acceleration (in part a) will be larger.

ANSWER

1) Choose sign convention.
Let $a$ be the magnitude of the acceleration with the sign convention:

for the block on the incline ( $M_{1}$ ) take positive toward the pulley (so the accelerations are consistent: if $M_{2}$ goes down, the rope pulls $M_{1}$ toward the pulley).
for the hanging block ( $M_{2}$ ) take positive downwards.
With this choice the two accelerations are the same scalar $a$ (opposite directions of their individual displacements).

Friction = $μ_{k} M_{1} g \cos θ$ . Whether the weight component $M_{1} g \sin θ$ is toward or away from the pulley depends on the slope orientation.

2) Draw free-body forces (along the directions we chose).

On $M_{1}$ (on the slope): tension $T$ acting up the slope, gravitational component $M_{1} g \sin θ$ acting down the slope, normal $N = M_{1} g \cos θ$ , kinetic friction $f_{k} = μ_{k} N = μ_{k} M_{1} g \cos θ$ acting opposite the motion (i.e. down the slope if $M_{1}$ tends to move up).
On $M_{2}$ (hanging): gravity $M_{2} g$ downward, tension $T$ upward.

3) Write Newton’s 2nd law for each mass (along the chosen axis).

Picture: the pulley sits at the lower end of the incline; moving toward the pulley means moving down the slope. Now the component $M_{1} g \sin θ$ acts toward the pulley.

For $M_{1}$ (positive toward the pulley = down the slope now):

Forces toward the pulley: $T$ (tension) and $M_{1} g \sin θ$ .
Friction opposes motion, so if $M_{1}$ moves toward the pulley friction acts away: $- μ_{k} M_{1} g \cos θ$ .

Newton for $M_{1}$ :

T + M_{1} g \sin θ - μ_{k} M_{1} g \cos θ = M_{1} a .

For $M_{2}$ (same as before):

M_{2} g - T = M_{2} a .

Add to eliminate $T$ :

M_{2} g + M_{1} g \sin θ - μ_{k} M_{1} g \cos θ = (M_{1} + M_{2}) a,

a = \frac{M_{2} g + M_{1} g \sin θ - μ_{k} M_{1} g \cos θ}{M_{1} + M_{2}} .

Now plug the numbers $M_{1} = 2 M, M_{2} = M, θ = 30^{\circ}, μ_{k} = 0.20$ :
Numerator $= M g + 2 M g \sin 30^{\circ} - 2 M g (0.20) \cos 30^{\circ}$

= M g (1 + 1 - 0.2 \cdot 2 \cdot 0.866) = M g (2 - 0.3464) = 1.6536 M g .

Divide by $3 M$ :

a \approx 0.5512 g \approx 0.5512 \times 9.8 \approx 5.40 {m/s}^{2} .

So in this orientation the hanging block falls (positive $a$ ), and the acceleration is much larger ( $\approx 5.40 {m/s}^{2}$ ).

Great — we’ll continue case 2 (pulley at the lower end, so the block $2 M$ slides down the slope toward the pulley and the hanging mass $M$ falls away from the pulley). I’ll (1) restate the acceleration from part (a) for this case, (2) give the distance–vs–time formula and a clear sketch description, and (3) give one modification that increases the acceleration (with a short proof).

Recap — acceleration

We derived

a = \frac{M_{2} g + M_{1} g \sin θ - μ_{k} M_{1} g \cos θ}{M_{1} + M_{2}} .

With $M_{1} = 2 M, M_{2} = M, θ = 30^{\circ}, μ_{k} = 0.20$ :

a = \frac{M g + 2 M g \sin 30^{\circ} - 2 M g (0.20) \cos 30^{\circ}}{3 M} = \frac{1 + 1 - 0.3464}{3} g = 0.5512 g \approx 5.40 {m/s}^{2} .

So the hanging block falls with constant acceleration $a \approx 5.40 m / s^{2}$ .

(b) Distance from the pulley vs time — equation and sketch

Let $s (t)$ be the distance of the suspended mass $M$ from the pulley, measured positive downwards (so $s$ increases as the block falls). Assume it starts at $t = 0$ from rest with initial distance $s_{0}$ and initial speed $v_{0} = 0$ . With constant acceleration $a$ ,

s (t) = s_{0} + v_{0} t + \frac{1}{2} a t^{2} = s_{0} + \frac{1}{2} a t^{2} .

With $a \approx 5.40 m / s^{2}$ ,

s (t) = s_{0} + 2.70 t^{2} (SI units) .

Key kinematic facts to mark on the sketch:

At $t = 0$ : $s = s_{0}$ and slope (instantaneous velocity) $= v (0) = 0$ .
The graph is a parabola opening upward (concave up) — distance increases increasingly fast because the block is accelerating downward.
Velocity vs time is linear: $v (t) = a t$ (straight line through origin).
Acceleration vs time is constant: $a (t) = 5.40 m / s^{2}$ (horizontal line).

If you want the time it takes to hit the pulley (i.e., to reduce the free length until the block $2 M$ hits the pulley), let the available fall distance be $Δ s$ (how far the hanging block can fall before the other block reaches the pulley). Then

t_{hit} = \sqrt{\frac{2 Δ s}{a}} .

(Example: if $Δ s = 1.0 m$ , $t_{hit} = \sqrt{2 / 5.40} \approx 0.61 s$ .)

(c) One modification to make the magnitude of the acceleration larger

Suggestion: Increase the incline angle $θ$ (make the slope steeper).

Why: In the formula for this orientation,

a = \frac{M_{2} g + M_{1} g \sin θ - μ_{k} M_{1} g \cos θ}{M_{1} + M_{2}},

increasing $θ$ increases $\sin θ$ and decreases $\cos θ$ . Differentiate the numerator with respect to $θ$ :

\frac{d}{d θ} (M_{1} g \sin θ - μ_{k} M_{1} g \cos θ) = M_{1} g (\cos θ + μ_{k} \sin θ) > 0

for $θ \geq 0$ and $μ_{k} \geq 0$ . So the numerator increases with $θ$ , hence $a$ increases.

Intuition: a steeper slope gives a larger component of $2 M$ ’s weight helping the motion and lowers the frictional resistance (because $\cos θ$ falls), so net accelerating force grows.

Other single changes that would also increase $| a |$ (if you prefer alternatives): increase the hanging mass $M_{2}$ , decrease the kinetic friction $μ_{k}$ (lubricate the incline), or reduce the mass $M_{1}$ on the slope. But the asked-for single modification above is increasing $θ$ .